The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4xx10^(10)s^(-1), calculate k at 318 K and E_(a).

The time required for 10% completion of a first order reaction at 298

1.	The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4xx10^(10)s^(-1), calculate k at 318 K and E_(a).
Answer» Solution :`k_(298"K")=(2.303)/(t_(1))LOG""(a)/(a-0.10a)=(2.303)/(t_(1))log""(10)/(9)=(2.303)/(t_(1))(0.0458)=(0.1055)/(t_(1))" or "t_(1)=(0.1055)/(k_(298"K"))` `k_(308"K")=(2.303)/(t_(2))log""(a)/(a-0.25a)=(2.303)/(t_(2))log""(4)/(3)=(2.303)/(t_(2))(0.125)=(0.2879)/(t_(2))" or "t_(2)=(0.2879)/(k_(308"K"))` But `t_(1)=t_(2)." Hence, "(0.1055)/(k_(298"K"))=(0.2879)/(k_(308"K"))" or "(k_(308"K"))/(k_(298"K"))=2.7289` Now, applying Arrhenius equation, `log""(k_(308"K"))/(k_(298"K"))=(E_(a))/(2.303"R")((T_(2)-T_(1))/(T_(1)T_(2)))` `:.log(2.7289)=(E_(a))/(2.303xx8.314" JK"^(-1)" mol"^(-1))xx((308-298)K)/(298" K"xx308" K")` `0.4360=(E_(a))/(2.303xx8.314)xx(10)/(298xx308)" or "E_(a)=76.623" kJ mol"^(-1).` CALCULATION of k at 318 K `log k=logA-(E_(a))/(2.303"RT")=log(4xx10^(10))-(7623" JK"^(-1)" mol"^(-1))/(2.303xx8.3174" JK"^(-1)"mol"^(-1)xx318" K")` `=10.6021-12.5843=-1.9822` or `k=" Antilog "(-1.9822)=" Antilog "(bar(2).0178)=1.042xx10^(-2)s^(-1).`